Is the set $U(n,\mathbb R)$ of all upper triangular $n\times n$ matrices over $\mathbb R$ a connected set in $M(n,\mathbb R)?$

Question

Is the set $U(n,\mathbb R)$ of all upper triangular $n\times n$ matrices over $\mathbb R$ a connected set in $M(n,\mathbb R)$ (with its usual topology after identification with $R^{n^2})?$

I think the answer is yes since connectedness is a productive property, $\mathbb R,\{0\}$ are connected and $$U(n,\mathbb R)=\\\mathbb R\times\mathbb R\times...\times\mathbb R\\\times\{0\}\times \mathbb R\times...\times\mathbb R\\...\\\times\{0\}\times\{0\}\times...\times\mathbb R$$

Please tell me whether the attempt is right or wrong!

Yes, it is right. Also, you could have constructed by hand the path from one upper triangular matrix to another upper triangular matrix. — Evgeny, Aug 09 '13 at 11:04

score 4 · Answer 1 · answered Aug 09 '13 at 11:04

4

If by "Productive" you mean "stable under product" then your reasoning is right.

You can also show it "by hand": for any two matrices in $U(n,\mathbb R)$, it is not hard to build a continuous path from one to the other.

answered Aug 09 '13 at 11:04

Denis

6,945
1
21
22

2

There is even a linear segment joining the two matrices, which proves that $U(n,\mathbb R)$ is convex. – lhf Aug 09 '13 at 11:07

score 2 · Answer 2 · answered Aug 09 '13 at 13:47

2

It is more, They are path connected too!

According to your notation, $A,B\in U(n,\mathbb{R})$, The continous path from $[0,1]$ is $f(t)=tA+(1-t)B$

answered Aug 09 '13 at 13:47

Myshkin

35,974
27
154
332

so they are convex too, isnt it? – vidyarthi Nov 30 '16 at 08:27
For convexity, we need a straightline path otherwise it is not convex, so yes they are convex. – Myshkin Mar 19 '19 at 10:22

Is the set $U(n,\mathbb R)$ of all upper triangular $n\times n$ matrices over $\mathbb R$ a connected set in $M(n,\mathbb R)?$

2 Answers2

Linked